A friend of mine told me that it's possible to hang a picture on the wall from a string using two nails in such a way that removing either of the two nails will make both the string and picture fall down. My friend also told me that I need to be acquainted with the concept of fundamental groups to understand the solution. The problem is that I'm not. Is there really no solution to this straightforward problem that doesn't require acquaintance with fundamental groups?
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asked May 8, 2014 at 11:30
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31$\begingroup$ I didn't realize until I started reading that there was a string or rope involved. All my pictures are hung by putting a nail or two into the wall, and then hanging the picture frame directly on the nail. In this case, with multiple nails, removing a nail always makes the picture fall, because with multiple nails, removing any nail causes the picture to be off balance, causing it to swing to one side, causing the picture to slide off the nail and crash to the floor. $\endgroup$– MichaelMay 8, 2014 at 16:42
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32$\begingroup$ @Michael. you sound like a real whiz around the house. Have you tried using one nail and hammering it through the picture into the wall - then it never falls off ? $\endgroup$– Tom CollingeMay 9, 2014 at 8:30
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3$\begingroup$ either of the two nails. $\endgroup$– TRiGMay 9, 2014 at 10:39
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7$\begingroup$ @Brilliand perhaps Tom was joking? ;) $\endgroup$– Cor_BlimeyMay 9, 2014 at 20:56
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2$\begingroup$ @Awesome this remind me of the joke where two nails walk into a bar... $\endgroup$– András HummerMay 10, 2014 at 13:21
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7 Answers
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You can see a solution in Figure 1 of Picture-Hanging Puzzles, by Demaine et al. It doesn't require anything advanced to understand that specific solution. It's the generalization to more and more nails that seems to need some fancy math.
Here is the two-nail solution from Demaine et al. (don't mouse over it if you want to think about it first):
An excerpt of an explanation found in section 3.2:
We define $2n$ symbols: $x_1, x_1^{-1}, \dots, x_n, x_n^{-1}$.
Each $x_i$ represents wrapping the rope around [passing over top of?] the $i$th nail clockwise, and each $x_i^{-1}$ represents wrapping the rope around the $i$th nail counterclockwise. Now a weaving of the rope can be represented by a sequence of these symbols. For example, the solution to the two-nail picture-hanging puzzle shown [in the figure above] can be written $x_1x_2x_1^{-1}x_2^{-1}$.
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In this representation, removing the $i$th nail corresponds to dropping all occurrences of $x_i$ and $x_i^{-1}$ in the sequence. Now we can see why [the figure] disentangles when we remove either nail. For example, removing the first nail leaves just $x_2x_2^{-1}$, i.e., turning clockwise around the second nail and then immediately undoing that by turning counterclockwise around the same nail. In general, $x_i$ and $x_i^{-1}$ cancel, so all occurrences of $x_ix_i^{-1}$ and $x_i^{-1}x_i$ can be dropped. (The free group specifies that these cancellations are all the cancellations that can be made.) Thus the original weaving $x_1x_2x_1^{-1}x_2^{-1}$ is nontrivially linked with the nails because nothing simplifies; but if we remove either nail, everything cancels and we are left with the empty sequence, which represents the trivial weaving that is not linked with the nails (i.e., the picture falls).
answered May 8, 2014 at 12:09
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10$\begingroup$ @Groo, thank you for adding the figure. $\endgroup$ May 8, 2014 at 14:15
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3$\begingroup$ Added a description of the solution from the paper. I recommend including figure 5 as a supplement to this, though I am uncertain of the precise extraction/licensing process for including it (though I'm sure Demaine et al. likely would not care). $\endgroup$– GankroMay 8, 2014 at 17:56
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1$\begingroup$ That sequence of operations is called a commutator. There's a nice Youtube video that gives a good explanation of the more general solution. $\endgroup$ Feb 26, 2020 at 13:21
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Put one nail in the wall, and another one in the picture frame. Wrap the string around them. If you remove any of the nails, the picture will fall down :)
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10$\begingroup$ Welcome to MSE. Good one :) $\endgroup$– user67773May 8, 2014 at 12:41
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How about puting two nails on the wall so that the picture is resting on top of the nails and there is no nail in the middle? Won't that do it?
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$\begingroup$ That would be supporting a picture from below, not hanging, I think? $\endgroup$ May 8, 2014 at 13:55
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2$\begingroup$ You can glue a piece of wood at the top horizontally and hang it from there... Anyhow, I am not serious about that, I knew there had to be a more sophisticated answer... $\endgroup$– GeorgyMay 8, 2014 at 15:02
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15$\begingroup$ @Georgy - This illustrates that the problem needs to be specified a little more formally. $\endgroup$– mbeckishMay 8, 2014 at 16:03
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$\begingroup$ How did you solve that mathematically? Can you give more details on the calculation you used to figure that out? $\endgroup$ May 9, 2014 at 8:16
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$\begingroup$ No, it was the first thing that came into my mind... $\endgroup$– GeorgyMay 9, 2014 at 13:27
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The solution can be understood without directly referring to the fundamental group (but knowing about fundamental groups makes it much easier to generalize to more nails).
Just consider wrapping the string once around one nail, once around the other and then the same again, but wrapping it around in the other direction (but still with the nails in the same order).
answered May 8, 2014 at 11:46
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Assuming the picture is a square, rotate it 45 degrees so it now has the shape of a "diamond" (rhombus). Now place a nail just under and a bit to the right of the left corner, and another nail just under and a bit to the left of the right corner.
answered May 8, 2014 at 13:04
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1$\begingroup$ For any rectangle there exists some angle and offsetting of nails that will work for this. I think nails halfway through each side, and the angle such that you can draw a vertical line through the top left corner and the bottom right corner should be close. $\endgroup$– CruncherMay 8, 2014 at 17:34
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Nail the picture on the center of left and right sides. The 'nail hold' of the frame have an opening perpendicular to the side, whose width equals the diameter of the nail. When one of nails are removed, the frame rotates till the nail reaches the opening. Once it have reached the opening, the nail will slipde through it and the picture will fall down.
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$\begingroup$ Ah, slots beside the nails. I knew there would be an easy solution ;) $\endgroup$– NavinMay 9, 2014 at 6:19
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Imagine having two nails in 3d, positioned perpendicular to the face of the wall (i.e., nails are horizontal), but with the two nails in perpendicular directions to each other and one nail above the other nail. It should be much easier to come up with a solution for how to wrap the string in this visualization so that the string falls when you remove either nail.. Then, once you have this solution, all you have to do is rotate one nail so that they both face the wall. This gives exactly the picture presented by Barry Cipra.
